Heat Calculation Considerations and Equations

Room heating calculations with factors for construction, outdoor temperatures, and other variables considered.

1. WALLS, ROOF, OR FLOORS WITH MINIMUM 3" INSULATION
   
2. WALLS, ROOF, OR FLOOR WITH 1" INSULATION
   
3. STUD WALL FINISHED BOTH SIDES - NO ADDED INSULATION
   
4. WOOD FLOORS
   
5. BRICK WALL - 8" THICK
   
6. WOOD SHEATHING ON STUDS
   
7. CEILING (NO INSULATION)
   
8. CORRUGATED METAL WALLS OR SINGLE GLASS WINDOWS

*Must also add wattage for air exchange - for 3 air exchanges per hour with 70� F temperature difference between room and outside - multiply cubic feet volume by 1.06 watts. For other temperature differences use equation:
Temperature Difference

-------------------------------------------------------------------------------- x 1.06 x CU.FT. Volume - see example below
70

Heat Calculation Example (use graph for approximate heat loss)

ROOM WITH FOLLOWING DIMENSIONS AND CONSTRUCTION FEATURES NEEDS TO BE HEATED TO 70� F WITH A WORSE CASE OUTSIDE TEMP OF 20� F

20 FT LONG X 10 FT WIDE X 8 FT
STUDDED WALLS FINISHED BOTH SIDES
NO WINDOWS/CEILING HAS NO INSULATION/WOOD FLOOR

A. OUTSIDE WALLS:
2(20 FT X 8 FT) + 2(10 FT X 8 FT) = 480 SQ. FT.
480 SQ. FT. X 5 W/SQ. FT. = 2400 WATTS (HEAT LOSS)

B. CEILING:
20 FT X 10 FT = 200 SQ. FT.
200 SQ. FT. X 11 W/SQ. FT. = 2200 WATTS (HEAT LOSS)

C. FLOOR:
20 FT X 10 FT = 200 SQ. FT.
200 SQ. FT. X 7 W/SQ. FT. = 1400 WATTS (HEAT LOSS)

D. ADDITIONAL HEAT FOR AIR EXCHANGES:
20 FT X 10 FT X 8 FT = 1600 CU. FT.
1600 CU. FT. X 50 F/70 F (RATIO) X 1.06 WATTS/CU. FT. = 1211 WATTS

RESULT: 2400 W (A) + 2200 W (B) + 1400 W (C) + 1211 W (D) + 1.2 (CONTINGENCY) 8640 W OR 8.64 KW
Room heat losses per �F - For various constructions

A. TO CALCULATE STRUCTURAL HEAT LOSSES:

1. DETERMINE AREA IN SQ. FT. FOR EACH SECTION OF WALL, CEILING, AND FLOOR
   
2. SELECT CORRESPONDING VALUE FOR SQ. FT. OF EACH SECTION AT LEFT SIDE OF GRAPH. MOVE FROM LEFT TO RIGHT ALONG HORIZONTAL LINE TO INTERSECTION POINT WITH CURVE FOR APPROPRIATE CONSTRUCTION
   
3. MOVE DOWN VERTICALLY TO HEAT LOSS VALUE (EXPRESSED IN WATTS PER DEG F)
   
4. REPEAT FOR 1-3 AS REQUIRED FOR ALL SECTIONS
   
5. ADD HEAT LOSS VALUES OF ALL SECTIONS TO DETERMINE TOTAL HEAT LOSS PER DEGREE FOR COMPLETE STRUCTURE
   
6. MULTIPLY TOTAL HEAT LOSS PER DEGREE BY DESIGN TEMPERATURE DIFFERENCE SET DESIRED INSIDE
   TEMPERATURES AND TYPICAL WORSE CASE OUTSIDE TEMPERATURE

B. TO CALCULATE AIR INFILTRATION HEAT LOSSES:

1. USE THE FOLLOWING VALUES FOR CALCULATION

NUMBER OF AIR EXCHANGES/HR WATT PER �F PER CU. FT.
1/4 .00129
1/2 .00258
3/4 .00387
1 .00516
2 .01032
3 .01548

2. CALCULATE VOLUME OF STRUCTURE = L X W X H (IN CUBIC FEET)
3. MULTIPLY CALCULATED VOLUME X WATT PER DEG F FOR AIR EXCHANGE
4. MULTIPLY RESULT FROM "STEP 3" BY DESIGN TEMPERATURE DIFFERENCE

C. ADD VALUES FOR "A" (HEAT LOSS) AND "B" (INFILTRATION HEAT LOSS)

RESULT = TOTAL HEAT LOSS FOR STRUCTURE